I'm talking about ONE manufacturer, giving a 50% wider FOV for two different "same numbers" binocs they make. So however they define FOV, it is THE SAME definition, from the same maker, for both binocs.
Same thing. I'm not comparing apples to oranges, I'm saying that it doesn't matter which glasses you compare. 7x50 to 7x50, or 8x42 to 8x42, pick ANY SIZE you want and then compare two different bincos IN THAT SAME NOMINAL SIZE. Whether they are made by the same maker or two different makers, but the SAME NUMBERS on both glasses. As I said, I'm familiar with optics, I know pretty well that a 7x and a 10x will--or at least should--have very different FOVs under most conditions.
I'm still not understanding this. First off, I'll assume when you say the length between the front and rear lens you mean the length of the optical path, since that can be folded differently. But even so, if the magnification is the same, and the size of the image coming out the back is the same (as measured by the exit pupil), then the distance shouldn't matter. Same is same, if the apparent view is 10x "bare eye" sized, you can only blow things up so far UNLESS you change the size of the image. So in theory a wider FOV would require a wider exit pupil image. That exit pupil image "crops" whatever the apparent FOV is going to be, doesn't it?
Doesn't make sense, unless the exit pupil becomes a meaningless number.
I think I see the assumptions that are leading you astray:
The exit lens doesn't crop the image, it just focusses the image for your eye. the far lens is what crops the image. For a wider FOV the outer lens would have to be closer to the pupil lens, or wider, while the pupil lens can stay the same.
Therefore if the far lens is closer to the pupil lens, the field of vision angle is short and _wide_, vs being farther away where the FOV angle gets longer and _narrow_, as the finger-loop demonstration shows.
What is being moved in the finger-loop demonstration is the far lens.
Does that help?